A Level H1 Biology Practice Paper 5

Answer Key - Biology H1 Practice Paper (Version 5)

Section A: Cell Structure and Biomolecules

Q1 (a) Phospholipids are arranged in a bilayer [1]; hydrophilic heads face the aqueous environments (extracellular and intracellular), while hydrophobic tails face inwards, away from water [1]. (b) The hydrophobic core prevents the free passage of polar/charged molecules and large molecules [1]. Only small, non-polar molecules can diffuse through [1]. This allows the cell to maintain internal concentrations different from the environment [1]. (c) Glucose moves via facilitated diffusion [1]. It passes through a specific carrier protein/channel [1] moving down its concentration gradient [1].

Q2 (a) Prokaryotes lack a membrane-bound nucleus; DNA is circular and free in cytoplasm [1]. Eukaryotes have a linear DNA enclosed in a nuclear envelope [1]. Prokaryotes lack membrane-bound organelles (e.g., mitochondria) [1], whereas eukaryotes possess them [1]. (b) The Golgi apparatus modifies, sorts, and packages proteins into vesicles [1]. If inhibited, proteins synthesized in the RER will not be processed or packaged correctly [1]. Consequently, secretory vesicles will not form or will contain non-functional proteins [1]. This leads to a failure in the secretion of proteins to the cell surface [1].

Q3 (a) The primary structure is the specific sequence of amino acids [1]. This sequence determines the R-group interactions (e.g., hydrogen bonds, ionic bonds, disulfide bridges, hydrophobic interactions) [1]. These interactions cause the polypeptide chain to fold into a specific 3D shape [1], which is the tertiary structure [1]. (b) Collagen consists of three polypeptide chains wound in a triple helix [1]. This structure provides immense strength and resistance to pulling forces [1]. The cross-linking between fibers further stabilizes the structure [1], making it ideal for connective tissues like tendons [1].

Q4 (a) Pyruvate can enter the mitochondrial matrix directly to be converted to acetyl-CoA and enter the Krebs cycle, where $\text{CO}_2$ is released [1]. Glucose cannot enter the mitochondria [1]; it must first undergo glycolysis in the cytoplasm to become pyruvate [1]. Isolated mitochondria lack the glycolytic enzymes needed to process glucose [1]. (b) The inner membrane is folded into cristae to increase the surface area [1]. This allows for more electron transport chain (ETC) proteins and ATP synthase molecules to be embedded [1]. A higher density of these proteins increases the rate of proton gradient formation and ATP synthesis [1].

Q5 (a) $\text{CO}_2$ must diffuse across the cell membrane into the chloroplast stroma for the Calvin cycle [1]. Water must enter root cells via osmosis across membranes to provide electrons via photolysis [1]. Active transport is required to move minerals (e.g., $\text{Mg}^{2+}$ for chlorophyll) against a gradient [1]. Glucose produced must be transported out of the chloroplast/cell via carrier proteins to be used elsewhere [1]. Without efficient membrane transport, the rate of photosynthesis would be limited by substrate availability [1].

Section B: Genetics and Inheritance

Q6 (a) DNA helicase unwinds the double helix [1]. DNA polymerase adds complementary nucleotides to the template strands [1]. Each original strand serves as a template for a new strand [1]. The result is two identical DNA molecules, each containing one old and one new strand [1]. (b) A point mutation may change a codon to specify a different amino acid (missense) [1]. This can alter the folding and active site of the protein [1]. Alternatively, a nonsense mutation creates a premature stop codon [1], leading to a truncated, non-functional protein [1].

Q7 (a) Parent 1: RR or Rr (Red); Parent 2: rr (White). Since F1 are all red, Parent 1 must be homozygous dominant (RR) [2]. (b) Parents: Rr x Rr. Gametes: R, r from both. Punnett square showing RR, Rr, Rr, rr [2]. Phenotypic ratio: 3 Red : 1 White [3].

Q8 (a) DNA is replicated exactly once during S-phase [1]. Sister chromatids are identical copies [1]. During anaphase, these chromatids are separated equally to opposite poles [1]. This ensures daughter cells are genetically identical to the parent cell [1]. (b) Mitosis produces two diploid daughter cells [1] that are genetically identical [1]. Meiosis produces four haploid daughter cells [1] that are genetically distinct due to crossing over and independent assortment [1].

Q9 (a) X-linked dominant inheritance [1]. (b) Affected fathers pass the X chromosome to all daughters [1], so all daughters are affected [1]. Fathers do not pass X to sons, so unaffected fathers cannot have affected daughters [1].

Q10 (a) Restriction enzymes cut DNA at specific recognition sites, creating "sticky" or "blunt" ends [2]. DNA ligase catalyzes the formation of phosphodiester bonds between the gene and the vector [2]. (b) Many bacteria reject foreign plasmids [1]. A selectable marker (e.g., antibiotic resistance gene) allows only those bacteria that have successfully taken up the plasmid to survive on a selective medium [2].

Section C: Energy and Physiology

Q11 (a) Competitive inhibitors bind to the active site of the enzyme [1]. They block the substrate from binding [1]. This increases the $K_m$ (lower affinity) but $V_{max}$ remains unchanged as high substrate concentration can displace the inhibitor [1]. (b) As temperature increases, kinetic energy increases, leading to more frequent collisions between enzyme and substrate [1]. At the optimal temperature, the rate is maximum [1]. Beyond this, heat disrupts hydrogen bonds/hydrophobic interactions [1], causing the enzyme to denature and lose its active site shape [1].

Q12 (a) APCs engulf pathogen and present antigens on MHC II [1]. Helper T-cells recognize antigen and release cytokines [1]. B-cells bind antigen and are activated by cytokines [1]. Activated B-cells undergo clonal expansion [1]. They differentiate into plasma cells [1], which secrete specific antibodies [1]. (b) Primary response is slower as B-cells must be activated and expanded [1]. Secondary response is much faster and stronger [1] because memory B-cells are already present and can rapidly differentiate into plasma cells [1].

Q13 (a) $\text{NAD}^+$ acts as an oxidizing agent/electron acceptor [1]. It is reduced to $\text{NADH}$ during the oxidation of glyceraldehyde-3-phosphate [1], allowing glycolysis to proceed [1]. (b) In anaerobic conditions, the ETC cannot oxidize $\text{NADH}$ [1]. Pyruvate is reduced to lactate by lactate dehydrogenase [1]. This process oxidizes $\text{NADH}$ back to $\text{NAD}^+$, allowing glycolysis to continue producing ATP [1].

Q14 (a) Variation exists in bacteria (e.g., via mutation) [1]. Some bacteria possess alleles for antibiotic resistance [1]. When antibiotics are applied, non-resistant bacteria die [1]. Resistant bacteria survive and reproduce [1], increasing the frequency of the resistance allele in the population [1]. (b) Allopatric speciation occurs when populations are geographically isolated [2]. Sympatric speciation occurs within the same geographic area, often due to behavioral or genetic isolation [2].

Q15 (a) mRNA binds to a ribosome [1]. tRNA molecules with anticodons complementary to mRNA codons bring specific amino acids [1]. The ribosome catalyzes peptide bond formation between amino acids [1]. The polypeptide chain grows until a stop codon is reached [1]. The chain then folds into its tertiary structure [1]. (b) Degenerate means multiple codons can code for the same amino acid [1]. This reduces the impact of some point mutations, as they may not change the resulting amino acid sequence [1].